Question

at noon, ship a is 170 km west of ship b. ship a is sailing east at 40 km/h and ship b is sailing north at 35 km/h. how fast (in km/hr) is the distance between the ships changing at 4:00 p.m.? (round your answer to three decimal places.)

Answer 1

The velocity of the distance between the ships at 4:00 p.m. is 53.144 km/h.

Step-by-step explanation:

To find the velocity of the distance between the ships at 4:00 p.m., we need to calculate the rates at which each ship is moving.

Ship A is sailing east at 40 km/h, so its velocity is 40 km/h.

Ship B is sailing north at 35 km/h, so its velocity is 35 km/h.

Since the two ships are moving at right angles to each other, we can use the Pythagorean theorem to find the velocity of the distance between them:

Velocity = √(40^2 + 35^2) = √(1600 + 1225) = √2825 = 53.144 km/h.

Therefore, the distance between the ships is changing at a rate of 53.144 km/h at 4:00 p.m.

Answer 2

The distance between ship A and ship B is changing at approximately 75.714 km/h at 4:00 p.m. This was determined by using the Pythagorean theorem to calculate the distance between the two ships as they move, and differentiating it with respect to time.

Step-by-step explanation:

To determine how fast the distance between the two ships is changing at 4:00 p.m., we can use the Pythagorean Theorem to represent the distance between them.

Let ship A have moved x km east from its original position after t hours, while ship B has moved y km north from its original position after t hours. At noon, the ships were 170 km apart, so when t = 0, x = 0 and y = 0.

Ship A's speed is 40 km/h east, and ship B's speed is 35 km/h north, so:

• x(t) = 40t
• y(t) = 35t

The distance d between the ships at any time t is given by:

d(t) = √((170 - x(t))^2 + y(t)^2) = √((170 - 40t)^2 + (35t)^2)

To find the rate at which distance d is changing with respect to time t, we need to differentiate d with respect to t:

Ωd/dt = (1/2) * [2*(170 - 40t)*(-40) + 2*(35t)*(35)] / √((170 - 40t)^2 + (35t)^2)

At 4:00 p.m., which is 4 hours after noon, we substitute t = 4 into the derivative to find the rate at which the distance is changing:

Ωd/dt at t = 4 = (1/2) * [-2*10*(-40) + 2*140*(35)] / √((10)^2 + (140)^2)

Ωd/dt at t = 4 = [800 + 9800] / √(19600)

Ωd/dt at t = 4 ≈ 10600 / 140 ≈ 75.714 km/h

The distance between the ships is changing at approximately 75.714 km/h at 4:00 p.m.